JEE Mains · Physics · STD 11 - 13. oscillations
A simple pendulum with length \(100\,cm\) and bob of mass \(250\,g\) is executing S.H.M. of amplitude \(10\,cm\). The maximum tension in the string is found to be \(\frac{x}{40}\,N\). The value of \(x\) is \(..........\).
- A \(98\)
- B \(97\)
- C \(99\)
- D \(100\)
Answer & Solution
Correct Answer
(C) \(99\)
Step-by-step Solution
Detailed explanation
\(\sin \theta_0=\frac{A}{l}=\frac{10}{100}=\frac{1}{10}\) From conservation of energy \(\frac{1}{2} m v^2=m g l(1-\cos \theta)\) Maximum tension occurs at mean position. \(\therefore T-m g=\frac{m v^2}{l}\) \(\Rightarrow T=m g+\frac{m v^2}{l}\)…
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